# Find the number of permutations and combinations if n = 12 and r = 2

• Last Updated : 30 Nov, 2021

Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

Permutation Formula

In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

nPr = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group

r = subset size, the number of things to be selected from the group

Combination

A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Combination Formula

In combination r things are picked from a set of n things and where the order of picking does not matter.

nCr = n!⁄((n-r)! r!)

Here,

n = Number of items in set

r = Number of things picked from the group

### Find the number of permutations and combinations if n = 12 and r = 2.

Solution:

Permutation Formula = nPr

nPr = n!/(n-r)!

= 12P2

= 12!/(12-2)!

= 12!/10!

= 12 × 11 × 10!/10!

= 132

Combination Formula = nCr

nCr = n!/(n-r)!r!

= 12C2

= 12!/(12-2)!2!

= 12!/10!2!

= 12×11×10!/10!2!

= 12×11/2

= 6 × 11

= 66

### Similar Questions

Question 1: Find the number of permutations and combinations if n = 10 and r = 2

Solution:

Permutation Formula = nP

nP = n!/(n-r)!

= 10P2

= 10!/(10-2)!

= 10!/8!

= 10×9×8!/8!

= 90

Combination Formula = nCr

nCr = n!/(n-r)!r!

= 10C2

= 10!/(10-2)!2!

= 10!/8!2!

= 10×9×8!/8!2!

= 10×9/2

= 5×9

= 45

Question 2: Find the number of permutations and combinations if n = 20 and r = 4

Solution:

Permutation Formula = nP

nP = n!/(n-r)!

= 20P4

= 20!/(20-4)!

= 20!/16!

= 20×19×18×17×16!/16!

= 116280

Combination Formula = nCr

nCr = n!/(n-r)!r!

= 20C4

= 20!/(20-4)!4!

= 20!/16!4!

= 20×19×18×17×16!/16!4!

= 20×19×18×17/4×3×2

= 5×19×3×17

= 4845

Question 3: Find the number of permutations and combinations if n = 22 and r = 8

Solution:

Permutation Formula = nP

nPr = n!/(n-r)!

= 22P8

= 22!/(22-8)!

= 22!/14!

= 22×21×20×19×18×17×16×15×14!/14!

= 12893126400

Combination Formula = nCr

nCr = n!/(n-r)!r!

= 22C8

= 22!/(22-8)!8!

= 22!/14!8!

= 22×21×20×19×18×17×16×15×14!/14!×8×7×6×5×4×3×2

= 22×21×20×19×18×17×16×15/8×7×6×5×4×3×2

= 11×5×19×3×17×2×3

= 319770

Question 4: Find the number of permutations and combinations if n = 8 and r = 2

Solution:

Permutation Formula = nP

nP= n!/(n-r)!

= 8P2

= 8!/(8-2)!

= 8!/6!

= 8×7×6!/6!

= 56

Combination Formula = nCr

nCr = n!/(n-r)!r!

= 8C2

= 8!/(8-2)!2!

= 8!/6!2!

= 8×7×6!/6!2!

= 8×7/2

= 4×7

= 28

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